The above volume maximization problem is simple to solve. What would its dual, the unfolded area minimization problem look like, given fixed volume of $V=\frac{8000}{27} \approx 296.3$? I would hope, since it is the dual, it would have the same solution dimensions:
- $h=x=\frac{10}{3}$,
- $l=\frac{40}{3}$, and
- $w=\frac{20}{3}$.
The problem is, I am having some trouble setting up and solving the unfolded area minimization problem. First attempt:
Objective: $\min \{(l+2x)(2w+2x)\}$ subject to $lwx=\frac{8000}{27}$.
We need to make some substitutions to make the objective function single-variable. One substitution we can make is $l+2x=2w+2x$, since it is assumed that the original unfolded cardboard is square. This gives
Objective: $\min \{(l+2x)(l+2x)\}$ subject to $lwx=\frac{8000}{27}$.
At this point, the objective function still has two variables, and it is not clear to me how to reduce the objective function further to a single variable.
Note 1: I believe the dual should be an area minimization (of unfolded, flat carboard), not surface area minimization (of folded box).
Note 2: I don't want to artificially make the problem simpler, e.g. I don't want to assume the folded box has a square base or anything like that. I only want to assume the cardboard is square when unfolded. Note that in the solution to the original problem, the folded box does NOT have a square base since the solution has $l \ne w \ne h$.
Note 3: This is intended to be solved by calc 1 (or even precalculus students, avoiding derivatives, by looking at a graph of the objective function). So please avoid solutions using derivatives or partial derivatives. My main objective here is to reduce the objective function to single-variable using the constraint that we are starting with a square sheet of cardboard.
Edit: After reading Vasili's comment, I lost confidence that I've specified the minimization problem correctly. We must assume (and be expected to use) the constraint that $lwx=\frac{8000}{27}$. And we also need to assume that the original piece of cardboard is square. Beyond that, I'm not sure if there's anything else that need be assumed for this minimization problem to be solvable. Could someone please double check that I have specified the correct dual problem?
Solution Following @Visili's helpful answer, I think we have a potential solution here. Using the fact that $l=2w$, we have $\frac{8000}{27}=lwx=2w^2x$. So $w^2=\frac{4000}{27x}$, or $w=\sqrt{\frac{4000}{27x}}$. Substituting this into the objective, we have \begin{align*} \min \left\{ A\right\} &=\min \left\{ (l+2x)(2w+2x)\right\} \\ &=\min \left\{(2w+2x)^2\right\} \\ &=\min \left\{\left(2\sqrt{\frac{4000}{27x}}+2x\right)^2\right\} \end{align*}
This gives minimum $\left(\frac{10}{3},400\right)$, which says the minimum cardboard area is $400$, attainable by selecting $x=\frac{10}{3}$, which was the result we hoped to see. Thanks @Visili - I think that solved the problem!
It's given that $lwx=\frac{8000}{3}$. Let $y$ be the side of the square. Thus, $l=y-2x, w=\frac{y}{2}-x \implies l=2w$.
The surface area: $A=2xl+2xw+2lw=4xw+2xw+4w^2$. Now use that $V=2w^2x$ to get a function of one variable.